Kaplan Combination Question

[nasir]

a company has 13 employees, 8 of whom belong to the union. if 5 people work any one shift and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift ?

total employees =13
# of unions employees =8
# of non union employees = 13-8= 5

we can select 4 employees from union workers = 8C4 = 8!/(8-4)!*4!
and 1 from non-union worker = 5C1 = 5!/(5-1)!*1!

=> 8C4 * 5C1 = 350

but the Answer explanation says that i have to add 8C5

8C5= the number of ways to select 5 union workers

i.e { 8C4 * 5C1) + 8C5 = 406 answer

i don't understand why we have to add 8C5.

Thanks

[diebeatsthegmat]

a company has 13 employees, 8 of whom belong to the union. if 5 people work any one shift and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift ?

total employees =13
# of unions employees =8
# of non union employees = 13-8= 5

we can select 4 employees from union workers = 8C4 = 8!/(8-4)!*4!
and 1 from non-union worker = 5C1 = 5!/(5-1)!*1!

=> 8C4 * 5C1 = 350

but the Answer explanation says that i have to add 8C5

8C5= the number of ways to select 5 union workers

i.e { 8C4 * 5C1) + 8C5 = 406 answer

i don't understand why we have to add 8C5.

Thanks

because we have to select AT LEAST 4 employees from union workers thus
1/ we select 4 employees from uni and 1 empoyee who is not from union
2/ we select totally 5 people from union
( because of the word AT LEAST we have to get 8C5)

[Maciek]

Nasir!
Your reasoning is correct.
8C5= the number of ways to select 5 union workers

The key word is 'at least'. Therefore, we need to add up combinations.

Either 4 union members work a shift or 5 union members work a shift.
Hence, we should add 8C5.

Hope it helps!
Best,
Maciek

[nasir]

i got it . Thank you guys !!